# I Area of sector by integration

1. Mar 5, 2016

### Noriele Cruz

I am bothered on how the area of circular sector was derived using integration. I have a solution but I don't know how it will go and come up with the formula:

A = 1/2 sr and A = 1/2 r2 θ

Can someone show me the solution for the derivation of the two formula?

2. Mar 5, 2016

### blue_leaf77

A basic question, what do you know about an integral?

3. Mar 5, 2016

### SteamKing

Staff Emeritus
It's an elementary proof if you use polar coordinates.

Examine the small sector in the figure below:

You can assume for a small angle dθ that the sector is roughly an isosceles triangle, where each side measures r units and the base is r ⋅ dθ units long.

The area of this triangle is dA ≈ (1/2) ⋅ r × r ⋅ dθ = r2 dθ / 2, using the formula for the area of a triangle.

If you want to find the area A of a circular sector between the angles θ1 and θ2, then

$$A = \int_{θ1}^{θ2} \,dA = \int_{θ1}^{θ2} \frac{1}{2} r^2\,dθ = \frac{1}{2} r^2 ⋅ (θ2 - θ1)$$

r ⋅ (θ2 - θ1) is also equivalent to the arc length s, and $A = \frac{1}{2} r ⋅ s$

Q.E.D.

4. Mar 5, 2016

### Noriele Cruz

I have done it by means of basic integration of plane areas. I divide the figure into two parts as A1 and A2 for computing the area.

It is somehow like this (I can't find a picture of my solution). I draw a triangle from A perpendicular to CB. Now, the sector is divided into a triangle and segment. The triangle represents A1 and segment represents A2. Can you show me your solution in my way of derivation?

5. Mar 5, 2016

### SteamKing

Staff Emeritus
No, because that's a really difficult way to do this problem.

There's nothing wrong with switching to integration using polar coordinates when that method makes the calculation much easier.

6. Mar 5, 2016

### Noriele Cruz

I am finish in integrating it with values.. I am stuck on how the integration will cancel out to come up with only 1/2 r2 θ for the area.

What do you mean by polar coordinates? I think I work with cartesian coordinates in integrating?

7. Mar 5, 2016

### SteamKing

Staff Emeritus
r is a constant with respect to theta. What happens when you integrate constants?
I mean polar coordinates:

https://en.wikipedia.org/wiki/Polar_coordinate_system

If you haven't studied polar coordinates, there is a certain mathematical background which is lacking in your education. You should correct this if you are going to study calculus.

8. Mar 5, 2016

### Noriele Cruz

I know how to integrate constants. I just don't know how to simplify my solution to 1/2 r2 θ.

We had study about polar coordinates and I know it, I just don't know how to make equations for the integration process.

9. Mar 5, 2016

### SteamKing

Staff Emeritus
You'll have to show your work where you got stuck.
I show you how and explained it in Post #3.

10. Mar 5, 2016

### Noriele Cruz

I got this part: πr2 / 4 - r2/2 arcsin (cos θ)

Can you show me how, please?

11. Mar 5, 2016

### Noriele Cruz

This is my solution:

#### Attached Files:

• ###### 12788044_1160874747263515_2108366211_n.jpg
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12. Mar 5, 2016

### SteamKing

Staff Emeritus
I'm not even sure what this expression is supposed to represent.

If it's supposed to be the area of a circular sector subtending an angle of 45°, it's way more complicated than it needs to be (and I haven't checked for any mistakes).

https://en.wikipedia.org/wiki/Circular_sector

13. Mar 5, 2016

### Noriele Cruz

Have you checked my solution in the attached file?

For the formula A = 1/2 sr, how it was derived using integration? Sorry for so many questions, I just want to clear up my mind.

14. Mar 5, 2016

### SteamKing

Staff Emeritus
I looked at it. It's way too complicated to spend time checking, especially when I have shown you in detail how to obtain a much simpler formula for the area of a circular sector, and showed you documentary proof of the validity of the simple formula.
Again, please refer to Post #3. I showed how this formula is derived from the alternate formula for the area of a circular sector.

If you are not going to read the posts I put up, there's no point in continuing this thread.

15. Mar 5, 2016

### Noriele Cruz

How come my picture of my integration of sector can be drawn like this? Is this also the same with mine?

16. Mar 6, 2016

### SteamKing

Staff Emeritus
All circular sectors look alike.

The dA in the figure is actually drawn for the case where r is not a constant but instead varies with the angle θ.

For a circular sector, r is constant, and integrating r and θ is not necessary to find the area of the sector.

17. Mar 6, 2016

### Noriele Cruz

How about the curve in the part of the sector? If I'm not mistaken, the figure you posted is somehow a triangle rather than a sector with curve part.
Where is it?

18. Mar 6, 2016

### SteamKing

Staff Emeritus
In the limit, as dθ → 0, then the arc portion of the sector is naturally going to appear more and more like a straight line segment, which is why the area of this tiny sector can be approximated using the formula for the area of a triangle.

To find the area of circular sector ABC, as shown in Post #4, that sector would be subdivided into many little slivers each sweeping out an angle dθ. The integration process adds up the areas dA of these segments as dθ → 0, producing a finite value A for the area of sector ABC.

19. Mar 7, 2016

### geoffrey159

In general, the area of a sector bounded by a curve $C$ defined by polar radius $\rho(\theta)$, such that $C = \rho([\theta_1,\theta_2])$ , is $A= \frac{1}{2} \int_{\theta_1}^{\theta_2} \rho^2(\theta) \ d\theta$ . In your case, $\rho(\theta)$ is constant and equal to $r$ since your sector is circular.

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